Swift - 将版本数组排序为字符串

Question

我有一个（应用程序版本的）字符串数组，随机排序如下：

var array = ["2.12.5", "2.12.10", "2.2", "2.11.8"]

排序后的数组应为 ["2.12.10"、"2.12.5"、"2.11.8"、"2.2"]（按最新排序）。我正在尝试对其进行排序。

//Compare by string
array.sort { //This returns ["2.2", "2.12.5", "2.12.10", "2.11.8"]
    $0 > $1
}
//Compare by int
array.sort { //This returns ["2.12.10", "2.12.5", "2.11.8", "2.2"]
    (Int($0.replacingOccurrences(of: ".", with: "")) ?? 0) > (Int($1.replacingOccurrences(of: ".", with: "")) ?? 0)
}

这些都无法正常工作。返回正确数组的最佳方法是什么？

Answer 1

最简单的选择是将compare与.numeric一起使用：

array.sort { $0.compare($1, options: .numeric) == .orderedDescending }

显然，如果您希望按升序排列，请使用 .orderedAscending .

Answer 2

您必须将版本字符串拆分为其组件，将它们转换为数字并比较它们的数值。试试这个：

var arrayOfStrings = ["2.12.5", "2.12.10", "2.2", "2.11.8"]
arrayOfStrings.sortInPlace {
    let v0 = ($0 as NSString).componentsSeparatedByString(".")
    let v1 = ($1 as NSString).componentsSeparatedByString(".")
    for i in 0..<min(v0.count, v1.count) {
        if v0[i] != v1[i] {
            return Int(v0[i]) < Int(v1[i])
        }
    }
    return v0.count < v1.count
}

这显然无法处理2.1a或3.14b123等版本号

Answer 3

Swift 4 版本：

var arrayOfStrings = ["2.12.5", "2.12.10", "2.2", "2.11.8"]
arrayOfStrings.sorted { (lhs, rhs) in
    let lhsc = lhs.components(separatedBy: ".")
    let rhsc = rhs.components(separatedBy: ".")
    for i in 0..<min(lhsc.count, rhsc.count) where lhsc[i] != rhsc[i] {
        return (Int(lhsc[i]) ?? 0) < (Int(rhsc[i]) ?? 0)
    }
    return lhsc.count < rhsc.count
}

Answer 4

Swift 4.2

：

var arrayOfStrings = ["1.2.1", "1.0.6", "1.1.10"]
arrayOfStrings.sort { (a, b) -> Bool in
    a.compare(b, options: String.CompareOptions.numeric, range: nil, locale: nil) == .orderedAscending
}
print(arrayOfStrings) // ["1.0.6", "1.1.10", "1.2.1"]